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If f : R → R is a function satisfying the equation f 3 x + 1 + f 3 x + 10 = 10 , ∀ x ∈ R , then the period of f x is

Correct answer

18

Step-by-step solution

∵ f 3 x + 1 + f 3 x + 10 = 10 … 1 Replacing x → x + 3 f 3 x + 3 + 1 + f 3 x + 3 + 10 = 10 f 3 x + 10 + f 3 x + 19 = 10 … 2 equation 2 - 1 f 3 x + 19 - f 3 x + 1 = 0 f 3 x + 1 = f 3 x + 19 Now, replace x → x 3 - 1 3 f 3 x 3 - 1 3 + 1 = f 3 x 3 - 1 3 + 19 f x = f x + 18     ∀ x ∈ R Hence, period of f x is 18

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